$20 \ mL$,$0.1 \ M \ HA$ $[(K_b)_{A^-} = 10^{-10}]$ is titrated against $0.2 \ M \ NaOH$. Select the correct statement [Given : $\log 2 = 0.3, \log 3 = 0.48$ ]

  • A
    Initially $pH$ of $HA$ solution is $11.5$
  • B
    At half equivalence point $pH$ of solution is $4$ and at equivalence point $pH$ is $9.09$
  • C
    Phenolphthalein is suitable indicator for this titration but methyl orange is not suitable
  • D
    Phenolphthalein and methyl orange both are suitable indicator for end point determination

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Similar Questions

When $100 \ mL$ of $1.0 \ M \ HCl$ was mixed with $100 \ mL$ of $1.0 \ M \ NaOH$ in an insulated beaker at constant pressure,a temperature increase of $5.7^{\circ} C$ was measured for the beaker and its contents (Expt. $1$). Because the enthalpy of neutralization of a strong acid with a strong base is a constant $\left(-57.0 \ kJ \ mol ^{-1}\right)$,this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. $2$),$100 \ mL$ of $2.0 \ M$ acetic acid $\left(K_a=2.0 \times 10^{-5}\right)$ was mixed with $100 \ mL$ of $1.0 \ M \ NaOH$ (under identical conditions to Expt. $1$) where a temperature rise of $5.6^{\circ} C$ was measured.
(Consider heat capacity of all solutions as $4.2 \ J \ g ^{-1} K ^{-1}$ and density of all solutions as $1.0 \ g \ mL ^{-1}$)
$1.$ Enthalpy of dissociation (in $kJ \ mol ^{-1}$) of acetic acid obtained from the Expt. $2$ is
$(A) \ 1.0 \ (B) \ 10.0 \ (C) \ 24.5 \ (D) \ 51.4$
$2.$ The $pH$ of the solution after Expt. $2$ is
$(A) \ 2.8 \ (B) \ 4.7 \ (C) \ 5.0 \ (D) \ 7.0$
Give the answer for question $1$ and $2.$

The concentration of $[H^{+}]$ and concentration of $[OH^{-}]$ of a $0.1 \ M$ aqueous solution of $2\%$ ionised weak acid is [Ionic product of water $= 1 \times 10^{-14}$]

At $90 \, ^\circ C$,the concentration of $H^+$ and $OH^-$ ions in pure water is $10^{-6} \, M$. What is the value of $[H^+] + [OH^-]$ at this temperature?

Statement $A$: Addition of $NH_4OH$ in the presence of excess $NH_4Cl$ to an aqueous solution of $BaCl_2$ results in the precipitation of $Ba(OH)_2$.
Reason $B$: $Ba(OH)_2$ is insoluble in water.

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The moles of $H^{+}$ from $H_2O$ alone in a $1 \ L$,$\sqrt{5} \times 10^{-7} \ M$ $HCl$ solution at $25 \ ^\circ C$ is ( $\sqrt{5} = 2.23$ )

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